Problem: [Æsir - CHAOS](https://soundcloud.com/kivawu/aesir-chaos)

[Æsir - V.](https://soundcloud.com/kivawu/aesir-v)

"Everything has been planned out. No more hidden concerns. The condition of Cytus is also perfect.

The time right now...... 00:01:12......

It's time."

The emotion samples are now sufficient. After almost 3 years, it's time for Ivy to awake her bonded sister, Vanessa.

The system inside A.R.C.'s Library core can be considered as an undirected graph with infinite number of processing nodes, numbered with all positive integers (1, 2, 3, …). The node with a number x (x > 1), is directly connected with a node with number (x)/(f(x)), with f(x) being the lowest prime divisor of x.

Vanessa's mind is divided into n fragments. Due to more than 500 years of coma, the fragments have been scattered: the i-th fragment is now located at the node with a number k_i! (a factorial of k_i).

To maximize the chance of successful awakening, Ivy decides to place the samples in a node P, so that the total length of paths from each fragment to P is smallest possible. If there are multiple fragments located at the same node, the path from that node to P needs to be counted multiple times.

In the world of zeros and ones, such a requirement is very simple for Ivy. Not longer than a second later, she has already figured out such a node.

But for a mere human like you, is this still possible?

For simplicity, please answer the minimal sum of paths' lengths from every fragment to the emotion samples' assembly node P.

Input

The first line contains an integer n (1 ≤ n ≤ 10^6) — number of fragments of Vanessa's mind.

The second line contains n integers: k_1, k_2, …, k_n (0 ≤ k_i ≤ 5000), denoting the nodes where fragments of Vanessa's mind are located: the i-th fragment is at the node with a number k_i!.

Output

Print a single integer, denoting the minimal sum of path from every fragment to the node with the emotion samples (a.k.a. node P).

As a reminder, if there are multiple fragments at the same node, the distance from that node to P needs to be counted multiple times as well.

Examples

Input


3
2 1 4


Output


5


Input


4
3 1 4 4


Output


6


Input


4
3 1 4 1


Output


6


Input


5
3 1 4 1 5


Output


11

Note

Considering the first 24 nodes of the system, the node network will look as follows (the nodes 1!, 2!, 3!, 4! are drawn bold):

<image>

For the first example, Ivy will place the emotion samples at the node 1. From here:

  * The distance from Vanessa's first fragment to the node 1 is 1. 
  * The distance from Vanessa's second fragment to the node 1 is 0. 
  * The distance from Vanessa's third fragment to the node 1 is 4. 



The total length is 5.

For the second example, the assembly node will be 6. From here:

  * The distance from Vanessa's first fragment to the node 6 is 0. 
  * The distance from Vanessa's second fragment to the node 6 is 2. 
  * The distance from Vanessa's third fragment to the node 6 is 2. 
  * The distance from Vanessa's fourth fragment to the node 6 is again 2. 



The total path length is 6.

Solution (in JAVA):

Solution:
/*
javac f.java && java f
*/

import java.io.*;
import java.util.*;

public class f {
	public static void main(String[] args) { new f(); }
	FS in = new FS();
	PrintWriter out = new PrintWriter(System.out);

	int n;
	long[] a;

	int MAX = 5001;
	int pcnt, pr[];
	int[][] vp, cs;
	boolean[] isnt;

	f() {
		pcnt = 0;
		isnt = new boolean[MAX];
		isnt[0] = true; isnt[1] = true;
		for (int i = 2; i < MAX; i++)
			if (!isnt[i]) {
				pcnt++;
				for (int j = (i << 1); j < MAX; j += i)
					isnt[j] = true;
			}
		pr = new int[pcnt]; pcnt = 0;
		for (int i = 0; i < MAX; i++)
			if (!isnt[i])
				pr[pcnt++] = i;

		/*
		long ops = 0, gap3 = 0;
		for (int i = 0; i < pcnt; i++) {
			long del = i == 0 ? 1 : pr[i] - pr[i - 1];
			out.printf("del = %d | p = %d%n", del, pr[i]);
			ops += (del * del * del);
			gap3 = max(gap3, del * del * del);
		}
		out.println("--> " + ops + " OW3 " + gap3);
		*/

		vp = new int[MAX][pcnt];
		for (int i = 2; i < MAX; i++)
			for (int j = 0; j < pcnt; j++) {
				int k = i;
				while (k > 0) {
					vp[i][j] += (k / pr[j]);
					k /= pr[j];
				}
			}
		cs = new int[MAX][pcnt + 1];
		for (int i = 2; i < MAX; i++)
			for (int j = 1; j <= pcnt; j++)
				cs[i][j] = vp[i][j - 1] + cs[i][j - 1];
		
		n = in.nextInt();
		a = new long[MAX];
		for (int i = 0; i < n; i++) {
			int x = in.nextInt();
			if (x == 0) x++;
			a[x]++;
		}
		long[] f = new long[MAX + 1];
		for (int i = 1; i <= MAX; i++)
			f[i] = f[i - 1] + a[i - 1];

		long[] dist = new long[MAX + 1];
		for (int i = 1; i <= MAX; i++)
			dist[i] = dist[i - 1] + a[i - 1] * cs[i - 1][pcnt];

		// F I X  T H I S
		long ans = dist[MAX];
		int l = 2, r; 
		for (int i = 0; i < pcnt; i++) {
			r = i == pcnt - 1 ? MAX - 1 : pr[i + 1] - 1;
			// everybody outside this range of highest prime must travel to 1 first
			// then travel up to the lca we're trying at each step
			long blw = dist[MAX] - (dist[r + 1] - dist[l]);
			// so when trying lca(j, k) we need to know how many primes are on in it
			// and how far every person in the range [l, r] will need to travel
			for (int x = r; x >= l; x--)
				for (int y = x; y >= l; y--) {
					// lca will be gotten as (pIDX, cnt)
					int[] lca = {-1, 0};
					for (int z = i; z >= 0; z--)
						if (vp[x][z] != vp[y][z]) {
							lca[0] = z;
							lca[1] = min(vp[x][z], vp[y][z]);
							break;
						}

					long up = lca[1] + cs[x][pcnt] - cs[x][lca[0] + 1];
					up *= (f[MAX] - f[r + 1] + f[l]);

					long abv = 0;
					for (int z = r; z >= l; z--) {
						if (x == z && y == z) continue;
						// given lca(x, y) = {w, cnt}
						// if we get to w add in prefix of other primes in a[z] * cs[w - 1]
						// add a[z] * [abs(cnt - cnt[z]) to abv
						// if we dont get to w before they differ
						// add a[z] * (diff to here + here up)
						for (int w = i; w >= lca[0]; w--) {
							if (w == lca[0]) {
								abv += a[z] * cs[z][w]; 
								abv += a[z] * abs(vp[z][w] - lca[1]);
								break;
							}
							if (vp[z][w] != vp[x][w]) {
								abv += a[z] * cs[z][w];
								abv += a[z] * abs(vp[z][w] - vp[x][w]);
								abv += a[z] * (cs[x][w] - cs[x][lca[0] + 1]);
								abv += a[z] * lca[1];
								break;
							}
						}
					}
					ans = min(ans, abv + blw + up);
				}
			l = r + 1;
		}

		out.println(ans);
		out.close();
	}
	
	int abs(int x) { if (x < 0) return -x; return x; }
	long abs(long x) { if (x < 0) return -x; return x; }
	int max(int x, int y) { if (x < y) return y; return x; }
	int min(int x, int y) { if (x > y) return y; return x; }
	long max(long x, long y) { if (x < y) return y; return x; }
	long min(long x, long y) { if (x > y) return y; return x; }
	int gcd(int x, int y) { while (y != 0) { x = y^(x^(y = x)); y %= x; } return x; }
	long gcd(long x, long y) { while (y != 0) { x = y^(x^(y = x)); y %= x; } return x; }
	long lcm(int x, int y) { long xy = x; xy *= y; return xy / gcd(x, y); }
	long lcm(long x, long y) { return (x * y) / gcd(x, y); }

	class FS {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens()) {
				try { st = new StringTokenizer(br.readLine()); }
				catch (Exception e) {}
			} return st.nextToken();
		}
		int nextInt() { return Integer.parseInt(next()); }
		long nextLong() { return Long.parseLong(next()); }
		double nextDouble() { return Double.parseDouble(next()); }

		void intArr(int sz, int[] x) { for (int i = 0; i < sz; i++) x[i] = nextInt(); }
		void longArr(int sz, long[] x) { for (int i = 0; i < sz; i++) x[i] = nextLong(); }
		void doubleArr(int sz, double[] x) { for (int i = 0; i < sz; i++) x[i] = nextDouble(); }
	}
}